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1 Answer
1

These notes written by Julien Melleray help us to solve the problem. I just state the results which will help in our case.

Lemma 3.3 (Pettis) Let $G$ a Polish group. For $A\subset G$, define
$U(A)$ as the biggest open set $V$ such that $A$ is comeagre $V$.
For any subsets $A$ and $B$ of $G$, we have
$$U(A)\cdot U(B)\subset A\cdot B.$$

As a consequence:

Theorem 3.4 Let $G$ a Polish group, and $A$ a Baire measurable non-meagre subset of $G$. Then $e$, the neutral element, belongs to the interior of $A\cdot A^{-1}$.

Back to the problem. Of course, $\Bbb R$ with the addition is a Polish group. Let $H$ a subgroup of $\Bbb R$ which is non-meagre and Borel measurable. It's Baire measurable. By the last theorem, $e$ belongs to the interior of $H\cdot H^{-1}=H$ as $H$ is a sub-group.

It's well-know that the subgroups of $\Bbb R$ are either of the form $a\Bbb Z$ (hence meagre) or dense. So we have a subgroup $H$ which is dense and has non-empty interior, say $(-r,r)$. Let $x\in \Bbb R$, and $x'\in H$ such that $|x-x'|\lt r$. Then $x-x'\in H$ and $x\in H$.

+1 for linking to Julien Melleray's notes, they look very nice at a first glance. I saw that you started with your PhD thesis: Good luck with it! Best wishes (t.b. from math.SE).
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Theo BuehlerNov 29 '12 at 21:31

Please keep in mind that the notes have not been carefully re-read and probably contain mistakes (and I'll be grateful, should you find any, if you let me know!)... Also, in the case of $\mathbb R$, probably the easiest way to answer the question is to use the Lebesgue measure, via the following statement (due to Steinhaus, I believe): if $A \subset \mathbb R$ is measurable of positive measure, then $A−A$ contains an interval centered around $0$ (this is a classical consequence of the regularity of the Lebesgue measure)
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Julien MellerayNov 29 '12 at 21:41

Thanks! I hope I will have luck with it, which will be the least we could expect of a probabilist!
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Davide GiraudoNov 29 '12 at 21:43

@Julien I thought about that, and it proves that a subgroup is either $\Bbb R$ or has $0$ measure. Is there a simple way to see it's meagre?
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Davide GiraudoNov 29 '12 at 21:45

2

A word of warning: Baire measurable is not the same as measurable with respect to the Baire $\sigma$-algebra (the $\sigma$-algebra generated by the compact $G_\delta$'s). It means the $\sigma$-algebra generated by the open sets and the meagre sets. There are meagre sets that aren't Borel: every set of reals can be written as the disjoint union of a meagre set $A$ and a Lebesgue null set $B$. If you partition a Vitali set this way: $V = A \cup B$ then $A$ can't even be Lebesgue measurable but is Baire measurable.
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Theo BuehlerNov 29 '12 at 21:46